LeetCode 114. 二叉树展开为链表

原题

LeetCode 114. 二叉树展开为链表

思路

• 这个先序遍历就是将右子树插入左子树的最右节点，再将左子树作为根节点的右子树。
• 第一种解法是看题解写的。确实清晰明了，而且不会错过任何一个情况。
• 第二种解法有两个值得注意的点。一是要记得将root->left设为nullptr，二是要记得将插入左子树最右节点的右子树也进行一个展开操作。

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode* root) {
        while (root != nullptr) {
            if (root->left == nullptr) root = root->right;
            else {
                TreeNode* left = root->left;
                while (left->right != nullptr) left = left->right;
                left->right = root->right;
                root->right = root->left;
                root->left = nullptr;
                root = root->right;
            }
        }
        return;
    }
};
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* flattenChildrenTree(TreeNode* root) {
        if (root == nullptr) return nullptr;
        TreeNode* right = flattenChildrenTree(root->right);
        root->right = flattenChildrenTree(root->left);
        TreeNode* tmp = root;
        while (tmp->right!= nullptr) tmp = tmp->right;
        tmp->right = right;
        root->left = nullptr;
        return root;
    }
    void flatten(TreeNode* root) {
        root = flattenChildrenTree(root);
        return;
    }
};
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运行截图

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